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Trigonometric form of complex numbers

Let M (x,y) be the point in the complex plane which is join to the complex number z = x + yi.

The position of the point M, and thus of the complex number z, we can determine by using numbers r and \varphi where r = |z| (the distance from the point M to the origin) and \varphi \in \left [0, 2\pi \right \rangle is an angle which the segment line \overline{OM} closing with the positive part of the real axis.

 

Then

    \[\cos \varphi = \frac{x}{r} \Rightarrow x = r \cos \varphi\]

and

    \[\sin \varphi = \frac{y}{r} \Rightarrow y = r \sin \varphi\]

is valid.

Substituting in the expression z=x + yi we obtain trigonometric form of  the complex number:

    \[z= r (\cos \vraphi + i \sin \varphi).\]

If  a complex number is given in the algebraic form z = x+yi, then r and \varphi we determine from equations:

    \[r = \sqrt{x^2 + y^2}\]

    \[tg \varphi = \frac{y}{x} , x \neq 0.\]

The last equation gives two solutions for an angle \varphi \in \left [0, 2\pi \right \rangle. We choose the angle depend on in which quadrant the number z is located. The angle \varphi is called an argument of the complex number z= x + yi and is denoted as arg(z).

Multiplication and division of complex numbers in trigonometric form

If z_1 = r_1 ( \cos \varphi_1 + i \sin \varphi_1) and z_2 = r_2 ( \cos \varphi_2 + i \sin \varphi_2), z_2 \neq 0, then their product is define as

    \[z_1 \cdot z_2 = r_1 r_2 ( \cos (\varphi_1 + \varphi_2) + i \sin (\varphi_1 + \varphi_2),\]

and their quotient as

    \[\frac{z_1}{z_2} = \frac{r_1}{r_2} ( \cos (\varphi_1 - \varphi_2) + i \sin (\varphi_1 - \varphi_2).\]

Example 1. Write in the trigonometric form complex numbers z and \overline{z} if z = \frac{1}{2} - \frac{\sqrt{3}}{2}i.

Solution:

We need to determine numbers r and \varphi.

    \[\begin{aligned} r &= |z| \\ &=|\frac{1}{2} - \frac{\sqrt{3}}{2}i| \\ &= \sqrt{\left( \frac{1}{2} \right) ^2 + \left( -\frac{\sqrt{3}}{2} \right)^2 } \\ &= \sqrt{\frac{1}{4}  + \frac{3}{4}} \\ & = \sqrt{1} \\ &= 1. \\ \end{aligned}\]

    \[tg \varphi = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3},\]

that is \varphi = \frac{5\pi}{3} or \varphi = \frac{8\pi}{3}. Since the number z = \frac{1}{2} - \frac{\sqrt{3}}{2}i is located in the fourth quadrant, it follows that \varphi = \frac{5\pi}{3}.

The complex number z= \frac{1}{2} - \frac{\sqrt{3}}{2} has the trigonometric form:

    \[z = \cos \frac{5 \pi}{3} + i \sin \frac{5\pi}{3}.\]

The complex conjugate numbers have the same modulus, therefore, for \overline{z} = \frac{1}{2} + \frac{\sqrt{3}}{2} , r =1. \overline{z} is located in the first quadrant, so we have:

    \[tg \varphi = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \Rightarrow \varphi = \frac{\pi}{3}.\]

Finally, the complex number \overline{z} = \frac{1}{2} + \frac{\sqrt{3}}{2} has the following trigonometric form:

    \[\overline{z} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}.\]

Example 2. Write in the trigonometric form the complex number z where:

    \[z = - \cos \frac{\pi}{5} + i \sin \frac{\pi}{5}.\]

Solution:

The function cosine is negative in the second and third quadrant, and sine is positive in the first and second quadrant. This means that the given complex number z is located in the second quadrant.

Now we have:

    \[tg \varphi = \frac{- \cos \frac{\pi}{5}}{\sin \frac{\pi}{5}} = - tg \frac{\pi}{5}.\]

That is, \varphi = - \frac{\pi}{5} or \varphi = \frac{4\pi}{5}.

We know that the complex number z is located in the second quadrant, which means that \varphi = \frac{4\pi}{5}. The radius of the same complex number is equal to 1.

Now we can write the given complex number z in the trigonometric form:

    \[z = \cos \frac{4 \pi}{5} + i \sin \frac{4 \pi}{3}.\]

 

De Moivre’s formula

For every complex number z  = r (\cos \varphi + i \sin \varphi) and n \in \mathbb{N}

    \[z^n = r^n ( \cos n \varphi + i \sin n \varphi)\]

is valid.

The formula above is called the De Moivre’s formula.

Example 3. Calculate

    \[\frac{(-1 + i)^{20}}{(\sqrt{2} - i \sqrt{2})^{10}}.\]

Solution:

Let  z_1 = -1 + i and z_2 = \sqrt{2} - i \sqrt{2}.

z_1 has the following trigonometric form:

    \[r_1 = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\]

    \[tg \varphi_1 = -1 \Rightarrow \varphi_1 = \frac{3\pi}{4},\]

because z_1 is in the second quadrant. Therefore

    \[z_1 = \sqrt{2} ( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}).\]

Now, by using the De Moivre’s formula, we have:

    \[\begin{aligned} z_1^{20} &= (-1 + i)^{20} \\ &=(\sqrt{2})^{20} \left ( \cos \left(20\cdot \frac{3\pi}{4}\right) + i \sin \left(20\cdot \frac{3\pi}{4}\right)\right) \\ &= 1024 ( \cos (15\pi) + i \sin (15\pi) \\ &= 1024 (\cos \pi + i \sin \pi) \\ \end{aligned}\]

For z_2 we have:

    \[r_2 = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2+2} = \sqrt{4} = 2\]

    \[tg \varphi_2  = -1 \Rightarrow \varphi_2  = \frac{7\pi}{4},\]

because the number z_2 is located in the fourth quadrant. It follows that the trigonometric form of the complex number z_2 is:

    \[z_2  = 2 ( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}).\]

 

By using the De Moivre’s formula we have:

    \[\begin{aligned} z_2^{10} &= (\sqrt{2} - i \sqrt{2})^{10} \\ &= 2^{10} \left ( \cos \left(10\cdot \frac{7\pi}{4}\right) + i \sin \left(10\cdot \frac{7\pi}{4}\right)\right) \\ & = 1024 \left ( \cos \left(\frac{35 \pi}{2}\right) + i \sin \left(\frac{35 \pi}{2}\right) \right) \\ & = 1024 \left ( \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) \right) \\ \end{aligned}\]

The rest is to calculate the quotient:

    \[\begin{aligned} \frac{z_1^{20}}{z_2^{10}} &= \frac{(-1 + i)^{20}}{(\sqrt{2} - i \sqrt{2})^{10}} \\ &=\frac{1024}{1024 } \left ( \cos \left ( \pi - \frac{\pi}{2} \right) + i \sin \left ( \pi - \frac{\pi}{2} \right) \right \\ & = 1 \cdot \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \\ & = i \end{aligned}\]

 

nth root of complex numbers

Every complex number z= r (\cos \varphi + i \sin \varphi) has exactly  n different values of n-th root given by the formula:

    \[\sqrt[n]{z} = \sqrt[n]{r} \left ( \cos \frac{\varphi + 2 k \pi}{n} + i \sin \frac{\varphi + 2 k \pi}{n} \right),\]

where k \in \{0, 1, 2, ..., n-1\} and n \in \mathbb{N}.

Example 4.  Determine \sqrt[6]{1}.

Solution:

The number z=1 + 0\cdot i has the following trigonometric form:

    \[r = \sqrt{1^2 + 0^2} = \sqrt{1} = 1\]

    \[tg \varphi = 0 \Rightarrow \varphi = 0\]

    \[\Rightarrow z = 1 (\cos 0 + i \sin 0).\]

Now we have, by using the formula for the nth root of a complex number for n=6:

    \[\sqrt[6]{1} = \sqrt[6]{1} \left ( \cos \frac{0 + 2k\pi}{5} + i \sin \frac{0 + 2 k \pei}{5} \right).\]

For k=0:

    \[z_0 = 1 (\cos 0 + i \sin 0) = 1.\]

For k = 1:

    \[z_1 = 1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \frac{1}{2} + i \frac{\sqrt{3}}{2}.\]

For k=2:

    \[z_2 = 1 \left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right) = - \frac{1}{2} + i \frac{\sqrt{3}}{2}.\]

For k=3:

    \[z_3 = 1 (\cos \pi + i \sin \pi) = -1.\]

For k=4:

    \[z_4 = 1 \left ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right ) = - \frac{1}{2} - i \frac{\sqrt{3}}{2}.\]

For k=5:

    \[z_5 = 1 \left ( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = \frac{1}{2} - i \frac{\sqrt{3}}{2}.\]

 

As we can observe, obtained complex numbers are arranged symmetrical. They form the vertices of a regular hexagon.

Similar, holds true in general. The formula for the nth root of complex numbers has a geometric interpretation. All nth roots of complex numbers have the same modulus \sqrt[n]{r}, which means that they are equally distant from the origin. Therefore, they all lie on the circle with center at the origin and radius \sqrt[n]{r}.

Furthermore, arguments of every two consecutive nth roots are differ for \frac{2 \pi}{n}, which means that these nth roots are arranged so that they form the vertices of a regular polygon.

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